Question

The loop below modifies bloc. Can sapply() achieve the same result?

Or in other words, can sapply() update variables in the base environment? I tried using assign() but without success. Thank you.

n <- 100 
fencePosts <- c(17,34) 
bloc <- rep(0,n) 
for (i in 1:length(fencePosts)){   
  bloc[fencePosts[i]:n] = i 
} 
table(bloc)

I was thinking of something like the following line, but of course bloc is out of scope (although I’m not sure why this doesn’t raise a “bloc not found” error).

zilch <- sapply(1:length(fencePosts),function(i)bloc[fencePosts[i]:n] = i)

Answer 1

I don’t know how kosher it is to the R-perts, but you could do

zilch <- sapply(1:length(fencePosts),function(i)bloc[fencePosts[i]:n] <<- i)